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Q.

A block of weight 5 N is pushed against a vertical wall by a force 12 N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is

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a

5 N

b

12 N

c

13 N

d

7.2 N

answer is D.

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Detailed Solution

$N = applied force = 12 N$

$∴ f_{\max} = μN = 7.2 N$

$Since, weight w < f_{\max}$

$Force of friction f = 5 N$

$∴ Net contact force = \sqrt{N^{2} + f^{2}} = \sqrt{{(12)}^{2} + {(5)}^{2}} = 13 N$

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A block of weight 5 N is pushed against a vertical wall by a force 12 N. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is